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I want to know whether $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^3+y^3}$ exists or not. I tried to approximate to (0,0) from different "paths" and the result was always 0. For example,

$f(x,mx^2) = \frac{m^2x^3}{1+m^3x^3}$

But that doesn't show that the limit is 0.

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The paths in my answer show that for any $\alpha$, there is a path so that $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y^2}{x^3+y^3}=\alpha$. Thus, the limit doesn't exist. – robjohn Oct 26 '12 at 5:53
up vote 6 down vote accepted

Hint: There is real trouble if $x$ is the negative of $y$. If you don't want to take the easy way out and note that the function is not defined when $x=-y\ne 0$, let $(x,y)$ approach $(0,0)$ along the parametric path $x=t+t^2$, $y=-t+t^2$. If you want more dramatic behaviour, use the path $(t+t^3, -t+t^3)$.

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The limite, for this path, is 1/6. But how did you come up with this? I mean, intuitively... – Thiago Oct 25 '12 at 0:45
@Thiago The idea is to be very close to $y=-x$. This suggest $x=t, y=-t$, but since the function is not defined there, you need to add something very small... If $t$ is small, then $t^2$ is much smaller ;) – N. S. Oct 25 '12 at 0:50
@Thiago You can try to understand this by drawing pictures for possible $(x, y) \in \mathbb{R}^{2}$. If there is a limit, then for any small neighborhood $D_{\epsilon}$ around $x^{2}y^{2}/(x^{3}+y^{3})$, you should be able to find some neighborhood $D_{\delta}$ around $(0,0)$ where every $(x, y) \mapsto x^{2}y^{2}/(x^{3} + y^{3}) \in D_{\epsilon}$. But even in that $D_{\delta}$, we have so many path to "screw" this quantity up. – user123454321 Oct 25 '12 at 0:57
@Thiago: N.S. has explained the motivation. The $t^2$ was (i) to be a tiny bit away from $y=-x$, but only a tiny bit; (ii) so $t^2$ is a natural shift to try, I tried and it worked; (iii) also, for fun, I wanted the limit along the path to be finite. – André Nicolas Oct 25 '12 at 0:58
@Thiago: More. The idea is that if we perturb $(t,-t)$ a little, then the "lead term" in $x^3+y^3$ will disappear, while the lead term in $x^2y^2$ won't. This will give us a good deal of control over the behaviour of the limit. Like you, I was changing to a one variable problem. Could have used $(t,-t+t^2)$, but have a liking for whatever symmetry might be around. Stuff like cancellation of lead terms happens a lot in one variable limits, particularly if we use series methods. – André Nicolas Oct 25 '12 at 2:04


Choose any $\alpha\ne0$. As $t\to-1$, consider the path $$ (x,y)=\alpha(t^3+1)\left(\frac1{t^2},\frac1{t}\right)\to(0,0) $$ For $\alpha=0$ use $$ (x,y)=(t^3+1,0)\to(0,0) $$ More Hint

For all $t\ne-1$, $\dfrac{x^2y^2}{x^3+y^3}=\alpha$.

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Hint: Use $\sqrt[3]{\frac{x^3+y^3}2}\ge \sqrt{xy}$.

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Only when $x,y \geq 0$. – user17762 Oct 25 '12 at 0:17

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